Question

Show that the force on each plate of a parallel plate capacitor has a magnitude equal to $\left(\frac{1}{2}\right)$ QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor .

Answer 1

Let $F$ be the force applied to separate the plates of a parallel plate capacitor by a distance of $△x$. Hence, work done by the force to do so $=F△x$

As a result, the potential energy of the capacitor increases by an amount given as $uA△x$

Where,

$u=$ Energy density

$A=$ Area of each plate

$d=$ Distance between the plates

$V=$ Potential difference across the plates

The work done will be equal to the increase in the potential energy i.e.,

$F△x=uA△x$

$F=uA=\left(\frac{1}{2}{\in }_0{E}^2\right)A$

Electric intensity is given by,

$E=\frac{V}{d}$

$\therefore F=\frac{1}{2}{\in }_0\left(\frac{V}{d}\right)EA=\frac{1}{2}\left({\in }_{0}A\frac{V}{d}\right)E$

However, capacitance, $C=\frac{{\in }_{0}A}{d}$

$\therefore F=\frac{1}{2}\left(CV\right)E$

Charge on the capacitor is given by,

$Q=CV$

$\therefore F=\frac{1}{2}QE$

The physical origin of the factor, $\frac{1}{2}$, in the force formula lies in the fact that just outside the conductor, field is E and inside it is zero. Hence, it is the average value, $\frac{E}{2}$, of the field that contributes to the force.